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Let $f \in W^{1,p}(U)$, then how to prove that $|f| \in W^{1,p}(U)$, where $W$ means the sobolev space over some open subset $U \in \mathbb{R}^n$.

In Lieb's Analysis he prove that Let $f$ be in $W^{1, p}(\Omega)$. Then the absolute value of $f$, denoted by $|f|$ and defined by $|f|(x)=|f(x)|$, is in $W^{1, p}(\Omega)$ with $\nabla|f|$ being the function $$ (\nabla|f|)(x)= \begin{cases}\frac{1}{|f|(x)}(R(x) \nabla R(x)+I(x) \nabla I(x)) & \text { if } f(x) \neq 0 \\ 0 & \text { if } f(x)=0\end{cases} $$ here $R(x),I(X)$ denote the real part and imaginary part of $f(x)$.

In the proof he uses the inequality $|\frac{1}{|f|(x)}(R(x) \nabla R(x)+I(x) \nabla I(x))|^2 \leq |\nabla R(x)|^2+|\nabla I(x)|^2$ and chain rule to show that $\nabla|f| \in L^p$, but how it implies $\partial |f| \in L^p(U)$, do we have $|\frac{1}{|f|(x)}(R(x) \partial R(x)+I(x) \partial I(x))|^2 \leq |\partial R(x)|^2+|\partial I(x)|^2$ ?

In all, I have three question:

  1. How to prove that $|\frac{1}{|f|(x)}(R(x) \nabla R(x)+I(x) \nabla I(x))|^2 \leq |\nabla R(x)|^2+|\nabla I(x)|^2$

  2. How to prove that $\partial |f| \in L^p(U)$.

  3. I wonder whether or not $|f| \in W_0^{1,p}(U)$ when $f \in W_0^{1,p}(U)$. I can't find a counterexample

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    $\begingroup$ The easy way to deal with absolute value of a function of Sobolev class, is approximating the function $x\to |f(x)|$ by $x\mapsto (f(x)^2+\epsilon^2)^{1/2}$, using the definition of the Sobolev space by weak derivatives, taking limits. $\endgroup$ Commented Nov 30, 2022 at 14:54

2 Answers 2

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Part 1 follows from the Cauchy-Schwartz inequality, applied to the two vectors $(R(x), I(x))$, $(\nabla R(x), \nabla I(x))$.

Part 2 follows from the simple inequality $|\partial_j R(x)| \leq \sqrt {\sum_i (\partial_i R(x))^2 } = |\nabla R(x)|$ for all $j$ (and similarly for $I$).

I am not sure about part 3 myself.

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  • $\begingroup$ thanks, it helps me a lot $\endgroup$
    – user494763
    Commented Dec 1, 2022 at 12:52
  • $\begingroup$ You're welcome! Do accept the answer if you're satisfied with it. $\endgroup$
    – Nate River
    Commented Dec 1, 2022 at 13:32
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Here is a result that proves part 3. It is copied from the paper:

P. Hajłasz, J. Maly, Approximation in Sobolev spaces of nonlinear expressions involving the gradient Ark. Mat. 40 (2002), 245--274.

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Let me explain the ``easy to see'' statement: $\min\{\varphi_k,u\}\to\min\{v,u\}$ in $W^{1,p}$.

Since $\min\{f,g\}=f-(f-g)^+$ ($+$ stands for the positive part), it suffices to prove that $$ f_k\to f \quad \Rightarrow \quad f_k^+\to f^+ \quad \text{in } W^{1,p}. $$ Inequality $|f_k^+-f^+|\leq |f_k-f|$ yields convergence in $L^p$ and convergence of the gradients follows from the fact that $$ \nabla f^+= \begin{cases} \nabla f & \text{if} f\geq 0,\\ 0 & \text{if } f<0. \end{cases} $$ This equality follows from the characterization of Sobolev functions though absolute continuity on lines, see Corollary 2.31 in

P. Hajlasz, Lecture Notes

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  • $\begingroup$ thanks, it's useful. But I dont understand why when the min function converges to u , then u is in the closure, Is the min already in the closure , or when take min on two smooth function we still get a smooth function? $\endgroup$
    – user494763
    Commented Dec 1, 2022 at 12:52
  • $\begingroup$ @user494763 I added some explanations. $\endgroup$ Commented Dec 1, 2022 at 14:31

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